# 2006 AIME A Problems/Problem 14

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## Problem

A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$)

## Solution $[asy] import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple AA = intersectionpoint(B--C, A--(-A) ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(B--D, black+1); draw(C--D, black+1); draw(A--AA, dashed); draw(B--O, dashed); draw(C--O, dashed); label("A", A, W); label("B", B, E); label("C", C, W); label("O", O, SE); label("D", D, Z); label("A'", AA, N); [/asy]$

From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \dfrac{BC\cdot \sqrt 3}2$, and therefore $BC=3\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\dfrac{BC\cdot AA'}2 = \dfrac{27\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\dfrac{ABC\cdot DO}3 = 9\sqrt 3$.

Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$:

import three; import math;
size(8cm,0);
currentprojection=obliqueX;

triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4);
triple BB=0.8*B + 0.2*D;
triple AC=(A+C)/2;

filldraw(A--BB--C--cycle, lightgray, black );

draw(A--B--C--cycle);
draw(O--D);
draw(A--D, black+1);
draw(BB--D, black+1);
draw(B--BB);
draw(C--D, black+1);
draw(A--O, dashed);
draw(B--AC, dashed);
draw(BB--AC, dashed);
draw(C--O, dashed);
label("$A$", A, W);
label("$B$", B, E);
label("$E$", BB, E);
label("$C$", C, W);
label("$O$", O, SE);
label("$D$", D, Z);
label("$B'$", AC, W);
(Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)')

The triangle $AEC$ will now be the base of the tripod, and our goal is to compute the height from $D$ onto the plane $AEC$.

First, consider the tetrahedron $ABCE$. It has the same base as $ABCD$, and as $E$ is in $1/5$ of $BD$, from similarity it follows that its height is $OD/5$, and therefore the volume of $ABCE$ is $1/5$ of the volume of $ABCD$.

But this means that we know the volume of $ACED$. All we have to do is to find the area of $ACE$, then we can easily compute the answer we need.

Set up a coordinate system where $O=(0,0,0)$, $B=(3,0,0)$, and $D=(0,0,4)$. Here we have $B'=(-3/2,0,0)$, and $E=(12/5,0,4/5)$.

Hence $EB' = \sqrt{ \left(\frac{12}{5} + \frac 32 \right)^2 + 0^2 + \left( \frac 45 \right)^2 } = \frac{\sqrt{1585}}{10}$.

We can now compute the area of $ACE$ as $\frac{AC\cdot EB'}2 = \frac{3\sqrt 3\cdot \sqrt{1585}}{20}$.

And finally, from the fact that the volume of $ACED$ is $4/5$ of the volume of $ABCD$ we have: \begin{align*} \frac{ACE \cdot x}3 & = \frac 45 \cdot 9\sqrt 3 \\ \frac{3\sqrt 3 \cdot \sqrt{1585} \cdot x}{3\cdot 20} & = \frac 45 \cdot 9\sqrt 3 \\ \frac{\sqrt{1585} \cdot x}{20} & = \frac 45 \cdot 9 \\ \sqrt{1585} \cdot x & = 16 \cdot 9 \\ x& = \frac{144}{\sqrt{1585}} \end{align*}

And as the prime factorization of $1585$ is $5\cdot 317$, we can conclude that $m=144$ and $n = 1585$.

As $40^2=1600 > 1585$ and $39^2 = 1521 < 1585$, we have $\lfloor \sqrt n\rfloor = 39$, and hence $\lfloor m+\sqrt n\rfloor = \boxed{183}$.